# Thread: slope of tangent to curve

1. ## slope of tangent to curve

find the slope of tangent to the curve x^2 +y^3=x-2 at the point (3,2)

so i want to get it in the form of f(x)=y

so y=(-x^2 +x-2)^1/3

so i then differentiated this using the chain rule and then i subbed in 3 for x and i thought i would get the right answer but no

2. ## Re: slope of tangent to curve

$y' = \dfrac{1}{3}\left(-x^2+x-2\right)^{-2/3}(-2x+1)$
$y'(3) = \dfrac{-5}{3(-8)^{2/3}} = \dfrac{-5}{12}$

Let's try by implicit differentiation:

$2x+3y^2\dfrac{dy}{dx} = 1$

$\dfrac{dy}{dx} = \dfrac{1-2x}{3y^2}$

Evaluate at $(3,2)$:

$\dfrac{dy}{dx} = \dfrac{1-2(3)}{3(2)^2} = \dfrac{-5}{12}$

I get the same answer from normal differentiation and implicit differentiation. What did you get?

3. ## Re: slope of tangent to curve

thanks. i was doing it wrong